009A Sample Final 3, Problem 7 - Revision history

Kayla Murray at 23:44, 2 December 2017

2017-12-02T23:44:51Z

2017-05-21T00:16:49Z

2017-05-20T23:43:35Z

2017-03-18T20:44:18Z

2017-03-07T19:38:45Z

2017-03-07T19:30:30Z

2017-03-07T19:26:28Z

2017-03-07T19:20:06Z

2017-03-06T21:32:17Z

2017-02-19T02:03:03Z

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp;<math style="vertical-align: -16px">\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 43: / Line 43: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9+x})}{(3+\sqrt{9+x})}}\\
+\displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9-x})}{(3+\sqrt{9-x})}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{9-(9+x)}}\\
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9-x})}{9-(9-x)}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-x}}\\
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9-x})}{x}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9+x}}{-1}}\\
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9-x}}{1}}\\
 &&\\
-& = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
+& = & \displaystyle{ \frac{3+\sqrt{9}}{1}}\\
 &&\\
-& = & \displaystyle{-\frac{6}{1}}\\
+& = & \displaystyle{\frac{6}{1}}\\
 &&\\
-& = & \displaystyle{-6.}
+& = & \displaystyle{6.}
 \end{array}</math>
 |-
@@ Line 120: / Line 120: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>-6</math>
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>6</math>
 |-
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|'''L'Hôpital's Rule'''
+|'''L'Hôpital's Rule, Part 1'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&nbsp; are both zero or both &nbsp;<math style="vertical-align: -1px">\pm \infty .</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;on an open interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; containing &nbsp;<math style="vertical-align: -5px">c,</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)\ne 0</math>&nbsp; on &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; except possibly at &nbsp;<math style="vertical-align: 0px">c.</math>&nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp; <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>
 |}

@@ Line 54: / Line 54: @@
 & = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
 &&\\
-& = & \displaystyle{\frac{6}{-1}}\\
+& = & \displaystyle{-\frac{6}{1}}\\
 &&\\
 & = & \displaystyle{-6.}
@@ Line 113: / Line 113: @@
 \displaystyle{\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\
 &&\\
-& = & \displaystyle{\frac{-5}{12}.}
+& = & \displaystyle{-\frac{5}{12}.}
 \end{array}</math>
 |}
@@ Line 125: / Line 125: @@
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>
 |-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math>\frac{-5}{12}</math>
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math>-\frac{5}{12}</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 29: / Line 29: @@
 !Step 1: &nbsp;
 |-
+|We begin by noticing that we plug in &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; into
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{x}{3-\sqrt{9-x}},</math>
 |-
+|we get &nbsp; <math style="vertical-align: -12px">\frac{0}{0}.</math>
 |}
@@ Line 41: / Line 39: @@
 !Step 2: &nbsp;
 |-
+|Now, we multiply the numerator and denominator by the conjugate of the denominator.
 |-
+|
@@ Line 105: / Line 121: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>-6</math>
 |-
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>

@@ Line 1: / Line 1: @@
 <span class="exam">Compute
-<span class="exam">(a) <math>\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}</math>
+<span class="exam">(a) &nbsp;<math style="vertical-align: -18px">\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}</math>
-<span class="exam">(b) <math>\lim_{x\rightarrow \pi} \frac{\sin x}{\pi-x}</math>
+<span class="exam">(b) &nbsp;<math style="vertical-align: -16px">\lim_{x\rightarrow \pi} \frac{\sin x}{\pi-x}</math>
-<span class="exam">(c) <math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math>
+<span class="exam">(c) &nbsp;<math style="vertical-align: -16px">\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"